Question: Divide the following complex numbers. $ \dfrac{-2+4i}{-3+i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3-i}$ $ \dfrac{-2+4i}{-3+i} = \dfrac{-2+4i}{-3+i} \cdot \dfrac{{-3-i}}{{-3-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-2+4i) \cdot (-3-i)} {(-3+i) \cdot (-3-i)} = \dfrac{(-2+4i) \cdot (-3-i)} {(-3)^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-2+4i) \cdot (-3-i)} {(-3)^2 - (1i)^2} = $ $ \dfrac{(-2+4i) \cdot (-3-i)} {9 + 1} = $ $ \dfrac{(-2+4i) \cdot (-3-i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2+4i}) \cdot ({-3-i})} {10} = $ $ \dfrac{{-2} \cdot {(-3)} + {4} \cdot {(-3) i} + {-2} \cdot {-1 i} + {4} \cdot {-1 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{6 - 12i + 2i - 4 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{6 - 12i + 2i + 4} {10} = \dfrac{10 - 10i} {10} = 1-i $